Prove using the definition on the limit that \lim_{n\to\infty}\frac{1+n}{1-2n}=\frac{-1}{2}

Prove using the definition on the limit that
$$\displaystyle\lim_{{{n}\to\infty}}{\frac{{{1}+{n}}}{{{1}-{2}{n}}}}={\frac{{-{1}}}{{{2}}}}$$

• Questions are typically answered in as fast as 30 minutes

Plainmath recommends

• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.

wheezym

$$\Bigg|\frac{1+n}{1-2n}+\frac{1}{2}\Bigg|$$
$$=\Bigg|\frac{2+2n+1-2n}{2(1-2n)}\Bigg|$$
$$=\Bigg|\frac{3}{2}\cdot\frac{1}{1-2n}\Bigg|$$
For every $$\displaystyle\epsilon{>}{0}$$ given we can choose N that
$$\Bigg|\frac{3}{2}\cdot\frac{1}{1-2}=\frac{3}{2(2N-1)}\Bigg|<\epsilon$$

$$\displaystyle{2}{N}-{1}{>}{\frac{{{3}}}{{{2}\epsilon}}}$$
$$\displaystyle{N}{>}{\left({1}+{\frac{{{3}}}{{{2}\epsilon}}}\right)}{\frac{{{1}}}{{{2}}}}$$
$$\displaystyle{N}{>}{\left({\frac{{{1}}}{{{2}}}}+{\frac{{{3}}}{{{4}\epsilon}}}\right)}$$
Hence $$\displaystyle\forall{n}\geq{N}$$
$$\Bigg|\frac{1+2n}{1-2n}-\bigg(-\frac{1}{2}\bigg)\Bigg|<\epsilon$$
Hence
$$\lim_{n\to\infty}\Bigg|\frac{1+2n}{1-2n}\Bigg|=-\frac{1}{2}$$